3 Marks Question

Question 13 Marks

A school has a rectangular play ground with length $x$ and breadth $y$ and a square lawn with side $x$ as shown in the figure given below. What is the total perimeter of both of them combined together?

Answer

Lenght of rectangular playground $AB = x$ And breadth of rectangular playground $BC = y$
$\because\ FCDE$ is a square i. e $FC = CD = EF = DE = x$
$\because\ ABCF$ is a rectangle i. e $AB = FC = x$ and $BC = AF = y [\because\ $All sides of a square are equal$]$

Now, perimeter of combined $($playground $+$ lawn$) =$ sum of all sides $= AB + BC + CD + DE + EF + FA = x + y + x + x + x + y =4x + 2y$

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Question 23 Marks

Amisha has a square plot of side $m$ and another triangular plot with base and height each equal to $m.$ What is the total area of both plots$?$

Answer

Given,
Side of square plot $=\mathrm{m}$ and height and base of triangular plot $=\mathrm{m}$
$\because$ Area of square plot $=\mathrm{m}^2$
$\left[\because\right.$ Area of square $\left.=(\text { side })^2\right]$
$\therefore$ Area of triangular plot $=\frac{1}{2} \times \mathrm{m} \times \mathrm{m}=\frac{\mathrm{m}^2}{2}$
$\left[\because\right.$ area of triangle $=\frac{1}{2} \times$ height $\times$ base $]$
$\therefore$ Totel of both $=$ area of square plot + area of tringular plot
$\mathrm{m}^2+\frac{\mathrm{m}^2}{2}=\frac{2 \mathrm{~m}^2+\mathrm{m}^2}{2}=\frac{3 \mathrm{~m}^2}{2}$
$[$Taking $LCM$ of $1$ and $2$ is $2 ]$

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Question 33 Marks

Arjun bought a rectangular plot with length $x$ and breadth $y$ and then sold a triangular part of it whose base is $y$ and height is $z.$ Find the area of the remaining part of the plot.

Answer

Given,
Arjun bought a rectangular plot with length $x$ and breasth $y.$
$\therefore\ $Area of rectangular plot $=$ Length $\times $ breadth $= x \times y = xy$
Alos given triangular part with base y and height z is sold
So, area of triangular part $=\frac{1}{2}\times\text{y}\times\text{z}=\frac{1}{2}\text{y}\ \text{z}$
$\Big[\therefore\ \text{area of triangle}\ = \frac{1}{2}\times\ \text{height}{\times}\text{base}\Big]$
$\therefore\ $Area of remainig part of the plot
$=$ Area of rectangular plot $-$ area of triangular plot
$=\text{xy}-\frac{1}{2}\text{xy}=\text{y}\Big(\text{x}-\frac{1}{2}\text{z}\Big)$

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Question 43 Marks

In the given figure, the length of a green side is given by $g$ and that of the red side is given by $p.$

Write an expression for the following pattern. Also write an expression if $100$ such shapes are joined together.

Answer

From the given figure Given, Lenght of green side $= g$ and length of red side $= p$ When we take three figures.

Total length of three figures $= 3(2g + 2p) = 6g + 6p$ If $100$ such shapes are joined together,
then the expression becomes. $= 100(2g +2p) = 200g + 200p = 200(g + p)$
Hence, the required expression is $200(g + p)$

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Question 53 Marks

What should be subtracted from $2 x^3-3 x^2 y+2 x y^2+3 y^3$ to get $x^3-2 x^2 y+3 x y^2+4 y^3$?

Answer

In order to get the solution we will subtract $x 3-2 x^2 y+3 x y^2+4 y^3$ from $2 x^3-3 x^2 y+2 x y^2+3 y^3$
Required expression is.
$ 2 x^3-3 x^2 y+2 x y^2+3 y^3-\left(x^3-2 x^2 y+3 x y^2+4 y^3\right)$
$=2 x^3-3 x^2 y+2 x y^2+3 y^3-x^3+2 x^2 y-3 x y^2-4 y^3$
On combining the like terms.
$=2 x^3-x^3-3 x^2 y+2 x^2 y+2 x y^2-3 x y^2+3 y^3-4 y^3$
$x^3-x^2 y-x y^2-y^3$
So, if we subrtact $x^3-x^2 y-y^3$ from $2 x^3-3 x^2 y+2 x y^2+3 y^3$
Then we get $x 3-2 x^2 y+3 x y^2+4 y^3$

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Question 63 Marks

What should be added to $x^3+3 x^2 y+3 x y^2+y^3$ to get $x^3+y^3$?

Answer

In order to find sobtract $x^3+3 x^2 y+3 x y^2+y^3$ from $x^3+y^3$
Required expression is
$x^3+y^3-\left(x^3+3 x^2 y+3 x y^2+y^3\right)$
$=x^3+y^3-x^3-3 x^2 y-3 x y^2-y^3$
On combining the like terms.
$=x^3-x^3+y^3-y^3-3 x^2 y-3 x y^2$
$=-3 x^2 y-3 x y^2$
So. if we add $-3 x^2 y-3 x y^2$ in $x^3+3 x^2 y+3 x y^2+y^3$
We get $x^3+y^3$

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Question 73 Marks

What should be subtracted from $-7 m n+2 m^2+3 n^2$ to get $m^2+2 m n+n^2$?

Answer

In order to get solution we will subtract $m^2+2 m n+n^2$ from $-7 m n+2 m^2+3 n^2$
Required expression is.
$-7 m n+2 m^2+3 n^2-\left(m^2+2 m n+n^2\right)$
$=-7 m n+2 m^2+3 n^2-m^2-2 m n-n^2$
On comibining the like terms.
$=-7 m n-2 m n+2 m^2-m^2+3 n^2-n^2$
$=-9 m n+m^2+2 n^2$
So, if we subract $m^2+2 n^2-9 m n$ from $-7 m n+2 m^2+3 n^2$
Then we get $m^2+2 m n+n^2$

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Question 83 Marks

What should be added to $3 p q+5 p^2 q^2+p^3$ to get $p^3+2 p^2 q^2+4 p q$?

Answer

In order to find solution sobtract $3 p q+5 p^2 q^2+p^3$ from $p^3+2 p^2 q^2+4 p q$
Required expression is
$p^3+2 p^2 q^2+4 p q-\left(3 p q+5 p^2 q^2+p^3\right)$
$=p^3+2 p^2 q^2+4 p q-3 p q-5 p^2 q^2-p^3$
On combining the like terms.
$=p^3-p^3+2 p^2 q^2-5 p^2 q^2+4 p q-3 p q$
$=-3 p^2 q^2+p q$
So. if we add $-3 p^2 q^2+p q$ in $3 p q^2+5 p^2 q^2+p^3$
We gwt $p^3+2 p^2 q^2+4 p q$

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Question 93 Marks

Add the following expression: $\frac{5}{8}\text{p}^4+2\text{p}^2+\frac{5}{8},\frac{1}{8}-17\text{p}\ +\frac{9}{8}\text{p}^2\ \text{and}\ \text{p}^5-\text{p}^3+7$

Answer

We have, $\Big(\frac{5}{8}\text{p}^4+2\text{p}^2+\frac{5}{8}\Big)+\Big(\frac{1}{8}-17\text{p}+ \frac{9}{8}\text{p}^2\Big)+(\text{p}^5 -\text{p}^3+7)$ $=\frac{5}{8}\text{p}^4+2\text{p}^2+\frac{5}{8}+\frac{1}{8}-17 \text{p}+\frac{9}{8}\text{p}^2+ \text{p}^5-\text{p}^3+7$ On combining the like terms. $= \text{p}^5+\frac{5}{8}\text{p}^4-\text{p}^3+\Big(2+\frac{9}{8}\Big)\text{p}^2-17\text{p}+\Big(\frac{5}{8}+\frac{1}{8}+7\Big)$ $= \text{p}^5+\frac{5}{8}\text{p}^4-\text{p}^3+\Big(\frac{16+8}{8}\Big)\text{p}^2-17\text{p}+\Big(\frac{5+ 1+56}{8}\Big)$ $=\text{p}^5+\frac{5}{8}\text{p}^4-\text{p}^3+\frac{25}{8}\text{p}^2-17\text{p}+\frac{62}{8}$ $=\text{p}^5+\frac{5}{8}\text{p}^4-\text{p}^3+\frac{25}{8}\text{p}^2-17\text{p}+\frac{31}{4}$

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Question 103 Marks

A triangle is made up of $2$ red sticks and $1$ blue sticks .

The length of a red stick is given by $r$ and that of a blue stick is given by $b.$ Using this information, write an expression for the total length of sticks in the pattern given below:

Answer

Given, length of a red stick a red $= 2r$ and renght of a blue stick $= b$ From the given figure the total number of red sticks $= 18$ and the total number of blue sticks $= 6$

So. the total lenght of sticks $= 18r + 6b = 6(3r + b)$
Hence.the required expression is $6(3r + b).$

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Question 113 Marks

In a rectangular plot, $5$ square flower beds of side $(x + 2)$ metres each have been laid (see figure given below). Find the total cost of fencing the flower beds at the cost of $Rs. 50$ per $100$ metres:

Answer

Given. Side of ane square flowe bed $= (x + 2)m$
$\therefore\ $ Perimeter of one square flowe bed $= 4 ($side$) = 4(x + 2)m$
Now, Totel perimeter of $5$ such square flower beds $= 5\ ×$ Perimere of one square $= 5 × 4(x + 2) = 20(x + 2)m$
$\because\ $Cost of fencing of $100m\ Rs. 50$
$\therefore\text{Cost of}\ 1\text{m}\ \text{Rs.}\frac{50}{100}$
$\therefore\ \text{Cost of}\ 20(\text{x}+2)\text{m}=\frac{50}{100}\times20(\text{x}+ 2)=10(\text{x}+2)$
$= \text{Rs.}(10\text{x} + 20)$

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Question 123 Marks

Subtract the sum of $12ab - 10 b^2-18 a^2$ and $9 a b+12 b^2+14 a^2$ from the sum of $a b+2 b^2$ and $3 b^2-a^2$.

Answer

$\text { Sum of } 12 a b-10 b^2-18 a^2 \text { and } 9 a b+12 b^2+14 a^2$
$=12 a b-10 b^2-18 a^2+9 a b+12 b^2+14 a^2$
On combining the like terms,
$=12 a b+9 a b-10 b^2+12 b^2-18 a^2+14 a^2$
$=21 a b+2 b^2-4 a^2$
Sum of $a b+2 b^2$ and $3 b^2-a^2$
$=a b+2 b^2+3 b^2-a^2$
$=a b+5 b^2-a^2$
Now, subtracting $21 a b+2 b^2-4 a^2$ from $a b+5 b^2-a^2$
we get:
$=\left(a b+5 b^2-a^2\right)-\left(21 a b+2 b^2-4 a^2\right)$
$=a b+5 b^2-a^2-21 a b-2 b^2+4 a^2$
$\text { On combining the like terms, }$
$=a b-21 a b+5 b^2-2 b^2-a^2+4 a^2$
$=-20 a b+3 b^2+3 a^2$
$=3 a^2+3 b^2-20 a b$

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Question 133 Marks

Observe the following nutritional chart carefully:

Food Item $($Per Unit $= 100g)$	Carbohydrates
Rajma	$60g$
Cabbage	$5g$
Potato	$22g$
Carrot	$11g$
Tomato	$4g$
Apples	$14g$

Write an algebraic expression for the amount of carbohydrates in $'g\ ’$ for.
$a. y$ units of potatoes and $2$ units of rajma.
$b. 2x$ units tomatoes and $y$ units apples.

Answer

$a.$ By unitary method,
$\therefore\ 1$ unit of potatoes contain carbohydrates $= 22g$
$y$ units of potatoes contain carbohydrates $= 22 \times y = 22yg$
Similarly,
$\therefore\ 1$ unit of rajma contain carbohydrates $= 60g$
$\therefore\ 2$ units of rajma contain carbohydrates $= (60 \times 2) = 120g$
Hence, the required expression is $22 y +120.$
$b.$ By unitary method,
$\therefore\ 1$ unit of tomatoes contain carbohydrates $= 4g$
$\therefore\ 2x$ units of tomatoes contain carbohydrates $= 2x \times 4 = 8xg$
Similarly,
$\therefore\ 1$ unit apples contain carbohydrates $= 14g$
$y$ units apples contain carbohydrates $= 14 \times y = 14yg$
Hence, the required expression is $8x +14y.$

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